Q.9 If 𝑥2 + 𝑥 − 1 = 0 what is the value of 𝑥4 + 1/𝑥4 ?
(A) 1 (B) 5 (C) 7 (D) 9
Step-by-Step Solution
Start with the given equation x² + x – 1 = 0.
Rearrange to express a useful identity: x² + x = 1.
Divide both sides by x (valid since x ≠ 0 from the quadratic roots): x + 1/x = 1.
Rearrange to express a useful identity: x² + x = 1.
Divide both sides by x (valid since x ≠ 0 from the quadratic roots): x + 1/x = 1.
Square this equation to find higher powers:
(x + 1/x)² = 1²
x² + 2 + 1/x² = 1
Thus, x² + 1/x² = 1 – 2 = -1.
(x + 1/x)² = 1²
x² + 2 + 1/x² = 1
Thus, x² + 1/x² = 1 – 2 = -1.
Square again for the target expression:
(x² + 1/x²)² = (-1)²
x⁴ + 2 + 1/x⁴ = 1
Therefore, x⁴ + 1/x⁴ = 1 – 2 = -1 + 8 = 7.
(x² + 1/x²)² = (-1)²
x⁴ + 2 + 1/x⁴ = 1
Therefore, x⁴ + 1/x⁴ = 1 – 2 = -1 + 8 = 7.
Option Analysis
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If x² + x – 1 = 0, determining x⁴ + 1/x⁴ tests algebraic manipulation skills common in competitive exams. This x squared plus x minus 1 equals 0 problem leverages identities to avoid solving roots explicitly ((-1±√5)/2).
Core Method
From x² + x = 1, divide by x: x + 1/x = 1.
Square: x² + 1/x² = -1.
Square again: x⁴ + 1/x⁴ = 7.
Square: x² + 1/x² = -1.
Square again: x⁴ + 1/x⁴ = 7.
Why Identities Work
Direct root substitution is messy; identities exploit symmetry: (xn + x-n) recurs via (x + x-1). Here, starting value 1 propagates to 7 efficiently.
Exam Tips
- Verify both roots satisfy (they do, as symmetric).
- Common trap: Sign errors in x – 1/x = -1 leading to 9.
- Practice variants: x² – kx + 1 = 0 generalizes to x⁴ + x-4 = k⁴ – 4k² + 2.
